POJ 1979 Red and Black （DFS的简单应用）

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 21765 Accepted: 11671

Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles. 

Write a program to count the number of black tiles which he can reach by repeating the moves described above. 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20. 

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows. 

'.' - a black tile 
'#' - a red tile 
'@' - a man on a black tile(appears exactly once in a data set) 
The end of the input is indicated by a line consisting of two zeros. 

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0

Sample Output

4559613

AC代码：

#include <stdio.h>#include <string.h>char mp[25][25];int vis[25][25],sum;void dfs(int x,int y){if(mp[x][y]=='#'||vis[x][y]==1)return ;sum++;vis[x][y]=1;dfs(x+1,y);dfs(x-1,y);dfs(x,y+1);dfs(x,y-1);}int main(){int w,h,i,j,x,y;while(~scanf("%d%d",&w,&h),(w+h)){getchar();sum=0;memset(vis,0,sizeof(vis));memset(mp,'#',sizeof(mp));for(i=1;i<=h;i++){for(j=1;j<=w;j++){scanf("%c",&mp[i][j]);if(mp[i][j]=='@'){x=i;y=j;}}getchar();}dfs(x,y);printf("%d\n",sum);}return 0;}